Let $M$ be a manifold of $L \to M$ a line bundle (say over $\mathbb{C}$, ie complex line bundle).
Is it true & why that for every such line bundle there exist a flat connection $\nabla_L : \Gamma(X,E)\to \Gamma(X, \Omega_X^1\otimes L)$, i.e. a connection which curvature $\nabla_L^2= \Omega_L \in \Omega ^{2}({\mathrm {End}}\,L)=\Gamma ({\mathrm {End}}\,L\otimes \Lambda ^{2}T^{*}M)$ is zero. Thus an existence problem. Sure, I not see any reason why all connections on $L$ should be flat, nevertheless I'm asking if on the other hand there always exist a flat one. If yes, is the claim independ of the field (so we can replace $\mathbb{C}$ by any other)?
Best Answer
No, this is false. Over $\mathbb{C}$ we have the exact sequence of sheaves of abelian groups on $X$ given by $$ 1 \to \mathbb{C}^\times \to \mathcal{O}_X^\times \stackrel{d \log}{\longrightarrow} \Omega_X \to 0. $$ The group $H^1(X, \mathbb{C}^\times)$ parametrizes line bundles with flat connection, and under this correspondence, the map to $H^1(X, \mathcal{O}_X^\times)$ (which parametrizes line bundles) forgets the connection. The obstruction to the surjectivity is $H^1(X, \Omega_X)$; a line bundle given by a transition function such that $d \log$ of the transition function is non-trivial in $H^1(X, \Omega_X)$ then gives a counterexample.